FindFunctionsWith-4CommonPoints

# FindFunctionsWith-4CommonPoints - From equation(4[a 3 =...

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In the Boolean Algebra S={0,1,2,3} and f:S 2 →S, we want to find BFs having the same values at these 4 points: f(2,0) = 3 ; f(2,3) = 0 ; f(1,3) = 3 ; f(3,3) = 1 Using expansion-2, the general expression is: f(x 1 ,x 2 ) = [a+x 1 +x 2 ][b+x 1 +x 2 ][c+x 1 +x 2 ][d+x 1 +x 2 ]. Using the 4 points: f(2,0)= [a+2+0][b+2+1][c+3+0][d+3+1] = [a+2][c+3] = 3 ………………. (1) f(2,3)= [a+2+3][b+2+2][c+3+3][d+3+2] = [b+2][c+3] = 0 ………………. (2) f(1,3)= [a+1+3][b+1+2][c+0+3][d+0+2] = [c+3][d+2] = 3 ………………. (3) f(3,3)= [a+3+3][b+3+2][c+2+3][d+2+2] = [a+3][d+2] = 1 ………………. (4)
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Unformatted text preview: From equation (4): [a+3] = 1 & [d+2] = 1 a = {2,1} & d = {3,1} ; but from (1), a=2 is not a solution: a=1. hence (4) and (1) are used. d = {3,1} and (3) gives c = {0,3}. c = {0,3} and (2) gives b = {0,2} Hence, solution: a=1, b={0,2}, c= {0,3}, d= {1,3} The corresponding function numbers: (1001) 4 =65 (1003) 4 =67 (1031) 4 =77 (1033) 4 =79 (1201) 4 =97 (1203) 4 =99 (1231) 4 =109 (1233) 4 =111 which verifies ryavuz2002.exe. results: USING CHOICE 3 and f(2,0)=3 ; f(2,3)=0 ; f(1,3)=3 ; f(3,3)=1 ; the same functions are obtained....
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## This note was uploaded on 10/18/2009 for the course EE ee431 taught by Professor Ee during the Spring '04 term at Istanbul Technical University.

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